How would you show that for any three loops $x, y, z$ that $(x*y)*z$ is equivalent to $x*(y*z)$. I want to show that $([x]*[y])*[z] = [x]*([y]*[z])$.
I am terrible at this stuff, would appreciate a good explanation and step by step proof or what I would need to do for proof. Thanks.
On
Relevant to this is my answer to On the associative property of a binary operation of the fundamental group. which shows you can define a path of length $r \geqslant 0$ from $x$ to $y$ to be a map $f: [0,r] \to X$ such that $f(0)=x, f(r) = y$. Composition of paths is then associative. If further $s \geqslant 0$ define $f +s$ to be $f$ on $[0,r]$ and constant on $[r,r+s]$. Define $f$ of length $r$ and $f'$ of length $r'$ to be equivalent if there are $s,s'$ such that $f+s, f'+s'$ are of the same length and are homotopic rel end points. You easily show any path is equivalent to a path of length $1$. In this way you get some easier proofs for the main properties of the fundamental group and groupoid.
Try to find a homeomorphism $p:[0,1]\to [0,1]$ such that $x*(y*z)=((x*y)*z)\circ p,$ and then use that any two paths $[0,1]\to [0,1]$ are homotopic.